Chữ gì mà nhỏ thế bn

\(\left(x^2+1\right)^2-4\left(2x-1\right)=0\)

\(\Leftrightarrow x^4+2x^2+1-8x+4=0\)

\(\Leftrightarrow x^4+2x^3-2x^3+x^2+5x^2-4x^2-10x+2x+5=0\)

\(\Leftrightarrow\left(x^4-2x^3+x^2\right)+\left(2x^3-4x^2+2x\right)+\left(5x^2-10x+5\right)=0\)

\(\Leftrightarrow x^2\left(x^2-2x+1\right)+2x\left(x^2-2x+1\right)+5\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow\left(x^2+2x+5\right)\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow\left(x^2+2x+5\right)\left(x-1\right)^2=0\)

Mà \(x^2+2x+5=\left(x+1\right)^2+4>0\)nên \(\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)

Vậy x = 1

\(ĐKXĐ:a,b,c\ne0\)

\(\frac{x-a}{bc}+\frac{x-b}{ca}+\frac{x-c}{ab}=\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\)

\(\Leftrightarrow\frac{xa-a^2}{abc}+\frac{xb-b^2}{abc}+\frac{xc-c^2}{abc}=\frac{2bc}{abc}+\frac{2ac}{abc}+\frac{2ab}{abc}\)

\(\Leftrightarrow\frac{xa-a^2+xb-b^2+xc-c^2}{abc}=\frac{2bc+2ac+2ab}{abc}\)

\(\Leftrightarrow xa-a^2+xb-b^2+xc-c^2=2bc+2ac+2ab\)

\(\Leftrightarrow xa+xb+xc=2bc+2ac+2ab+a^2+b^2+c^2\)

\(\Leftrightarrow x\left(a+b+c\right)=\left(a+b+c\right)^2\)

\(\Leftrightarrow x=a+b+c\)

Vậy x = a + b + c

\(ĐKXĐ:a,b,c\ne0\)

\(\frac{a+b-x}{c}+\frac{b+c-x}{a}+\frac{c+a-x}{b}+\frac{4x}{a+b+c}=1\)

\(\Leftrightarrow\frac{a+b-x}{c}+\frac{b+c-x}{a}+\frac{c+a-x}{b}=1-\frac{4x}{a+b+c}\)

\(\Leftrightarrow1+\frac{a+b-x}{c}+1+\frac{b+c-x}{a}+1+\frac{c+a-x}{b}=4\)

\(-\frac{4x}{a+b+c}\)

\(\Leftrightarrow\frac{a+b+c-x}{c}+\frac{a+b+c-x}{a}+\frac{a+b+c-x}{b}=\)

\(\frac{4\left(a+b+c\right)}{a+b+c}-\frac{4x}{a+b+c}\)

\(\Leftrightarrow\frac{a+b+c-x}{c}+\frac{a+b+c-x}{a}+\frac{a+b+c-x}{b}=\)

\(\frac{4\left(a+b+c-x\right)}{a+b+c}\)

\(\Leftrightarrow\left(a+b+c-x\right)\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}-\frac{4}{a+b+c}\right)=0\)

\(\Rightarrow\left(a+b+c-x\right)=0\)hoặc \(\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}-\frac{4}{a+b+c}\right)=0\)

+) Nếu \(\Rightarrow\left(a+b+c-x\right)=0\)thì x = a + b + c

+) Nếu \(\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}-\frac{4}{a+b+c}\right)=0\)thì x thỏa mãn với mọi số

\(\frac{x}{2000}+\frac{x+2}{2002}+\frac{x+4}{2004}+....+\frac{x+12}{2012}=7\)

\(\Leftrightarrow\left(\frac{x}{2000}-1\right)+\left(\frac{x+2}{2002}-1\right)+\left(\frac{x+4}{2004}-1\right)+......+\left(\frac{x+12}{2012}-1\right)=0\)

\(\Leftrightarrow\frac{x-2000}{2000}+\frac{x-2000}{2002}+\frac{x-2000}{2004}+.....+\frac{x-2000}{2012}=0\)

\(\Leftrightarrow\left(x-2000\right)\left(\frac{1}{2000}+\frac{1}{2002}+\frac{1}{2004}+....+\frac{1}{2012}\right)=0\)

Dễ thấy \(\frac{1}{2000}+\frac{1}{2002}+....+\frac{1}{2012}>0\Rightarrow x-2000=0\Rightarrow x=2000\)

\(\frac{x+1}{15}+\frac{x+2}{7}+\frac{x+4}{4}+6=0\)

\(\Leftrightarrow\left(\frac{x+1}{15}+1\right)+\left(\frac{x+2}{7}+2\right)+\left(\frac{x+4}{4}+3\right)=0\)

\(\Leftrightarrow\frac{x+16}{15}+\frac{x+16}{7}+\frac{x+16}{4}=0\)

\(\Leftrightarrow\left(x+16\right)\left(\frac{1}{15}+\frac{1}{7}+\frac{1}{4}\right)=0\)

Dễ thấy \(\frac{1}{4}+\frac{1}{7}+\frac{1}{15}>0\Rightarrow x+16=0\Rightarrow x=-16\)

Ta có :

\(\frac{3x+15}{x^2-5x+4}=\frac{A}{x-1}+\frac{B}{x-4}\)

\(\Leftrightarrow\frac{3x+15}{\left(x-4\right)\left(x-1\right)}=\frac{A\left(x-4\right)+B\left(x-1\right)}{\left(x-1\right)\left(x-4\right)}\)

\(\Leftrightarrow3x+15=Ax-4A+Bx-B\)

\(\Leftrightarrow3x+15=\left(A+B\right)x-\left(4A+B\right)\)

\(\Leftrightarrow\hept{\begin{cases}A+B=3\\4A+B=-15\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}A=-6\\B=9\end{cases}}\)

\(\Leftrightarrow A-B=-15\)

Vậy A - B = 15

P/s :>>  Câu này có trong Vio Toán Tv cấp trường 8 năm 2019-2020

=> Tớ vừa thii xong :) Ở bài tăng dần đầu tiên :vv

Chết .... Làm đúng mà KL sai bét :)

Sửa lại hộ tớ là -15 nhé !!

a) Gọi diện tích cả cánh đồng là a (a > 0)

Ta có phương trình :

\(a-\left(\frac{7}{18}a+\frac{4}{15}a\right)=31\)

\(\Leftrightarrow\frac{31}{90}a=31\)

\(\Leftrightarrow a=90\)

Vậy diện tích cánh đồng là 90ha

b) Diện tích phần trồng khoai là :

        \(\frac{7}{18}a=\frac{7}{15}.90=35ha\)

Vậy diện tích trồng khoai 35 ha

Trl :

-Câu này có trong Vio Toán tv lớp 8 ( tớ vừa mới thi ạ :33 )

-Hơi ngại làm :> Nhưng cho cậu kq nhé : 162 cm²

100%

Cảm ơn chú đã kb giờ thì t sẽ làm hộ chú :V

\(P=\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}+\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}-1\right):\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}-\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}+1\right)\)

\(P=\left[\frac{\left(\sqrt{a}+1\right)\left(\sqrt{ab}-1\right)}{\left(\sqrt{ab}+1\right)\left(\sqrt{ab}-1\right)}+\frac{\left(\sqrt{ab}+\sqrt{a}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}-\frac{ab-1}{ab-1}\right]\)

\(:\left[\frac{\left(\sqrt{a}+1\right)\left(\sqrt{ab}-1\right)}{\left(\sqrt{ab}+1\right)\left(\sqrt{ab-1}\right)}\right]-\frac{\left(\sqrt{ab}+\sqrt{a}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}+\frac{ab-1}{ab-1}\)

\(P=\frac{\left(a\sqrt{b}-\sqrt{a}+\sqrt{ab}-1\right)+\left(ab+\sqrt{ab}+a\sqrt{b}+\sqrt{a}\right)-\left(ab-1\right)}{ab-1}\)

\(:\frac{\left(a\sqrt{b}-\sqrt{a}+\sqrt{ab}-1\right)-\left(ab+\sqrt{ab}+a\sqrt{b}+\sqrt{a}\right)+\left(ab-1\right)}{ab-1}\)

\(P=\frac{a\sqrt{b}-\sqrt{a}+\sqrt{ab}-1+ab+\sqrt{ab}+a\sqrt{b}+\sqrt{a}-ab+1}{ab-1}\)

\(:\frac{a\sqrt{b}-\sqrt{a}+\sqrt{ab}-1-ab-\sqrt{ab}-a\sqrt{b}-\sqrt{a}+ab-1}{ab-1}\)

\(P=\frac{2a\sqrt{b}+2\sqrt{ab}}{ab-1}:\frac{-2\sqrt{a}-2}{ab-1}\)

\(P=\frac{2\sqrt{ab}\left(\sqrt{a}+1\right)}{ab-1}.\frac{ab-1}{-2\left(\sqrt{a}+1\right)}=-\sqrt{ab}\)

P/s: :V bài này tính toán kĩ nhưng chưa chắc đúng :VVVV

Mr.Fuff cảm ơn ạ >_<!!!!

hello

1 cách cực ngu 

\(\frac{x+107}{7}+\frac{x+6}{47}+\frac{x+1}{33}+\frac{x+184}{21}=0\)

\(\Leftrightarrow1551\left(x+107\right)+231\left(x+6\right)+329\left(x+1\right)+517\left(x+184\right)=0\)

\(\Leftrightarrow1551x+165957+231x+1386+329x+517x+95128=0\)

\(\Leftrightarrow2628x+262800=0\)

\(\Leftrightarrow2628x=-262800\)

\(\Leftrightarrow x=-100\)

Đặt \(x^2+x=t\)

Khi đó phương trình tương đương với:

\(t-\frac{4}{t}=3\)

\(\Rightarrow t^2-3t-4=0\)

\(\Leftrightarrow\left(t^2-1\right)-\left(3t+3\right)=0\)

\(\Leftrightarrow\left(t-1\right)\left(t+1\right)-3\left(t+1\right)=0\)

\(\Leftrightarrow\left(t+1\right)\left(t-4\right)=0\)

\(\Rightarrow t=-1;t=4\)

Thay vào làm nốt